Question about commutative diagram

Question

We consider the following diagram of $R$-modules, when $R$ is a commutative ring with unit. \begin{alignat}{3} 0 \xrightarrow{} &M &\xrightarrow{\;u\;} &\;N &\xrightarrow{\;v\;} &\; Q \rightarrow 0 \\ \llap{f}&\downarrow &\llap{g}&\downarrow &\llap{h}&\downarrow \\ 0 \rightarrow &M'& \xrightarrow{u'} &\;N' &\xrightarrow{v'}&\;Q' \rightarrow 0 \end{alignat} If the maps $ f: M \xrightarrow{} M'$ and $h : Q \xrightarrow{} Q'$ are isomorphisms and the map $g : N \xrightarrow{} N'$ is a monomorphism. We suppose also that the first line is exact and $\operatorname{Im}(u') \subseteq \operatorname{Ker}(v')$. If the diagram is also commutative i.e. $u'f=gu$ and $v'g=hv$. Under these assumptions, is it possible that the second line is not exact, i.e. there exists $n' \in \operatorname{Ker}(v') \backslash \operatorname{Im}(u')?$

Answer 1

Yes, there is an example of this. Take $M=M'=\mathbb Z=N=N'$ and $Q=Q'=\mathbb Z/2\mathbb Z$ all as $\mathbb Z$-modules. For the first row take as $u$ multiplication by $2$ and as $v$ the projection onto the quotient. For the second row take as $u$ multiplication by $6$ and as $v'$ again the projection. Finally, take as $g$ multiplication by $3$.

As $3\equiv1\mod2$ the diagram commutes, the first row is exact but the second row is not since $4\notin\operatorname{im}(u')$ but $4\in\ker(v')$. Hence $\operatorname{im}(u')\subsetneq\ker(v')$.